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How do you integrate #int e^(2x)/(1+e^(2x))dx#?

1 Answer

Narad T. · mason m

Nov 21, 2016

The answer is #=1/2ln(1+e^(2x))+C#

Explanation:

Let's do it by substitution

Let #u=1+e^(2x)#

then, #du=2e^(2x)dx#

#int(e^(2x)dx)/(1+e^(2x))#

#=1/2int(du)/u=1/2lnabsu#

#=1/2ln(1+e^(2x))+C#

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